Let $f(x, y) = 4ye^x$. Suppose $\vec{a} = (1, -1)$ and $\vec{v} = \left( -1, 0 \right)$. Find the directional derivative of $f(x, y)$ at $\vec{a}$ in the direction of $\vec{v}$. $\dfrac{\partial f}{\partial v} = $
Answer: When a directional derivative is in the direction $(1, 0)$, $(0, 1)$, $(-1, 0)$, or $(0, -1)$, it becomes a regular partial derivative. Because $v = (-1, 0)$, the directional derivative we want to find is also $-\dfrac{\partial f}{\partial x}$ evaluated at $(1, -1)$. $\begin{aligned} &-\dfrac{\partial f}{\partial x} = -4ye^x \\ \\ &-\dfrac{\partial f}{\partial x}(1, -1) = 4e \end{aligned}$ In conclusion, the directional derivative of $f$ at $\vec{a}$ in the direction of $\vec{v}$ equals $4e$.